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CUSTOMER_CODE SMUDE DIVISION_CODE SMUDE EVENT_CODE APR2016 ASSESSMENT_CODE BC0039_APR2016 QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 5592 QUESTION_TEXT Define i.Union sets ii.Intersection sets iii.Difference sets iv.Complement of a set SCHEME OF EVALUATION i.Union Sets: The union of two sets A and B denoted by AUB is the set of elements which belongs to A or B or both.That is, AUB={x:x∈A or x∈B} ii.Intersection of sets: The intersection of two sets A and B denoted by A∩B is the set of elements which belongs to both A and B.That is, A∩B={x:x∈A and x∈B} iv.Difference of sets: The difference of two sets A and B, denoted by A–B is the set of elements of A which are not the elements of B. That is, A–B={x:x∈A, x∈B} v.Complement of a set: The complement of a set A with respect to the universal set U is defined as U–Aand is denoted by A` or A c. That is A`={x:x∈U, x∉A} QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 5593 QUESTION_TEXT Let R be the set of real numbers. Define f:R→R by i.f(x)=2x+3 ii.f(x)=x 3 for every x in R, prove that f is one–one? i.f(x)=2x+3 Solution: Let f(x 1)=f(x 2) for some x 1, x 2∈R 2x 1+3=2x 2+3 X 1=x 2 SCHEME OF EVALUATION Thus for every x 1, x 2∈R, f(x 1)=f(x 2) implies x 1=x 2. Therefore f is one–one. ii.f(x)=x 3 Solution: Let f(x 1)=f(x 2) for some x 1,x 2∈R = X 1=x 2 Therefore f is one–one QUESTION_T DESCRIPTIVE_QUESTION YPE QUESTION_ID 73263 QUESTION_T a. EXT b. Show that [p (p q)] is a tautology. Verify that [p (p q)] is a contradiction. a.) Now we write down the truth table Observing the table we can conclude that [p (p q)] is always true. Hence [p SCHEME OF (p q)] is a tautology. EVALUATION b. Solution: Now we write down the truth table Observing the table, we can conclude that [p (p q)] false. Hence [p (p q)] is always is a contradiction. QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 105218 QUESTION_TEXT Prove that if H is a non empty finite subset of a group G and H is closed under multiplication, then H is a subgroup of G. Proof: Suppose H is a non empty finite subset of a group G and H is closed under multiplication SCHEME OF EVALUATION Now, we have to show that H is a subgroup of G. It is enough to show that a ∈H ⟹ a–1 ∈H. Since, H is a non empty, there exists a ∈ H. Now, a∈ H ⟹ a2 ∈H. Similarly a3 ∈H, …… am ∈H, …….. Therefore H≥ {a, a2, …….}. Since H is finite, we name that there must be repetitions in a, a2….. Therefore there exists integers r, s with r>s > 0 Such that ar = as ⟹ ar. a–s = a0 ⟹ ar–s = e ⟹ e ∈H (since r–s >0, a∈ H ⟹ar–s ∈H) Since r–s–1 ≥0, we have ar–s–1 ∈H and a. ar–s–1 = ar–s = e ∈H. Therefore ar–s–1 acts as the inverse of a in H. Hence H is a subgroup of G QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 105224 Prove that if V be a vector space over F and W is a nonempty subset of V. Then the following conditions are equivalent. QUESTION_TEXT i. W is a subspace of V ii. α, β ∈ F and W1, W2 ∈ W implies αW1 + βW2 ∈ W. (i) ⇒ (ii): α ∈ F, w1 ∈ W ⇒ αw1 ∈ W; β ∈ F, w2 ∈ W ⇒ βw2 ∈ W. SCHEME OF EVALUATION Since αw1, βw2 ∈ W and W is subspace we have αw1 + βw2 ∈ W. (ii) ⇒ (i): Let w1, w2 ∈ W. Since 1 ∈ F we have that w1 + w2 = 1.w1 + 1. w2 = αw1 + βw2 ∈ W (here α = 1 and β = 1). Therefore + is a closure operation on W. Since W ⊆ V, the associative law holds. x ∈ W ⇒ x, x ∈ W ⇒ 0 = 1.x + (1) = αx + βx ∈ W (here α = 1 and β = -1). Therefore the additive identity 0 is in W. Let x ∈ W. Then x, x ∈ W ⇒ -x = 0. x + (-1)x = αx + βx (here α = 1 and β = -1). Therefore the additive inverse –x ∈ W for any element x ∈ W. Hence (W, +) is a group. Since W ⊆ V and V is Abelian, we have that (W, +) is also Abelian. Also it is easy to verify that the other conditions of vector space. Therefore W is a subspace of V. QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 105226 QUESTION_TEXT Define the following with one example. Set, Subset, cardinality of a set, Universal set Set: A well defined collection of objects is called a set. Ex: B= {1, 3, 5, 7, 9} Subset: If every element of a set A is also an element of a Set B, then A is said to be a sub set of B. Ex: N is the Set of Natural no’s: SCHEME OF EVALUATION Z is the Set of integers. Then NCZ (N is a Subset of Z) Cardinality of a set: If A is a finite set, then the Cardinality of A is the total number of elements that comprise the set and denoted by N(A). Ex: A = {1, 2, 3, 4, 5} Then n(A) = 5 Universal Set: If all the Sets are subsets of a fixed Set, then this Set is called the universal set & denoted by U Ex: In the Study of theory of numbers the set Z of integers is considered as the universal set.